Question

Which of the following compounds reacts with${\mathrm{NaNO}}_2$and $\mathrm{HCl}$ at $-4°\mathrm{C}$ to yield alcohol (or) phenol?

• A. ${\mathrm{C}}_2{\mathrm{H}}_5{\mathrm{NH}}_2$
• B. ${\mathrm{C}}_6{\mathrm{H}}_5{\mathrm{NH}}_2$
• C. ${\mathrm{CH}}_3{\mathrm{NHCH}}_3$
• D. ${\mathrm{C}}_6{\mathrm{H}}_5{\mathrm{NHCH}}_3$

Answer 1

The correct option is A

${\mathrm{C}}_2{\mathrm{H}}_5{\mathrm{NH}}_2$

The explanation for the correct option

(A) ${\mathbf{C}}_{\mathbf{2}}{\mathbf{H}}_{\mathbf{5}}{\mathbf{NH}}_{\mathbf{2}}$

• ${\mathrm{C}}_2{\mathrm{H}}_5{\mathrm{NH}}_2$ (ethylamine) is a primary amine so can react with ${\mathrm{NaNO}}_2$ (sodium nitrite)with $\mathrm{HCl}$ (hydrochloric acid) to form a diazonium salt.
• This diazonium salt on further treatment with water can form alcohol.
• The reaction is as follows,
• ${\mathrm{NaNO}}_2+\mathrm{HCl}\stackrel{-4^0\mathrm{C}}{\to }{\mathrm{HNO}}_2+\mathrm{NaCl}$
• $$\begin{gathered} {\mathrm{C}}_2{\mathrm{H}}_5{\mathrm{NH}}_2+{\mathrm{HNO}}_2\stackrel{}{\to }{\mathrm{C}}_2{\mathrm{H}}_5-\stackrel{+}{\mathrm{N}}\equiv \mathrm{N} \\ \mathrm{diazonium}\mathrm{salt} \end{gathered}$$
• $$\begin{gathered} {\mathrm{C}}_2{\mathrm{H}}_5-\stackrel{+}{\mathrm{N}}\equiv \mathrm{N}\stackrel{{\mathrm{H}}_2\mathrm{O}}{\to }{\mathrm{C}}_2{\mathrm{H}}_5\mathrm{OH}+{\mathrm{N}}_2 \\ \mathrm{ethanol} \end{gathered}$$
• Thus option A is correct

The explanation for the incorrect options

(B)${\mathbf{C}}_{\mathbf{6}}{\mathbf{H}}_{\mathbf{5}}{\mathbf{NH}}_{\mathbf{2}}$

• Aniline${\mathrm{C}}_6{\mathrm{H}}_5{\mathrm{NH}}_2$ is also a primary amine but it is aromatic.
• only primary aliphatic amine responds to ${\mathrm{NaNO}}_2$and $\mathrm{HCl}$to form diazonium salt.
• Thus option B is incorrect

(C) ${\mathrm{CH}}_3{\mathrm{NHCH}}_3$

• Dimethylamine ${\mathrm{CH}}_3{\mathrm{NHCH}}_3$is a secondary amine. Thus cannot form alcohol.
• It will form respective nitrosamine compound as follows,

• Thus option C is incorrect

(D) ${\mathrm{C}}_6{\mathrm{H}}_5{\mathrm{NHCH}}_3$

• N-methylaniline is a secondary amine.
• Thus it will give a nitroso compound as follows,

• Thus option D is incorrect

Therefore, the correct option is (A) ${\mathbf{C}}_{\mathbf{2}}{\mathbf{H}}_{\mathbf{5}}{\mathbf{NH}}_{\mathbf{2}}$ .